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Mice21 [21]
3 years ago
9

A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the

butcher can sell to the next customer
Physics
1 answer:
KiRa [710]3 years ago
8 0

Answer:

The maximum amount of meat that the butcher can sell is  3\frac{1}{12}\:lb

Explanation:

The maximum amount can be found by taking the difference of mixed numbers.

5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\

Best Regards!

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What is kinematics?<br>explain!!~​<br><br>thankyou ~
Hatshy [7]

Answer:

<u>Kinematics:-</u>

  • Branch of mechanics concerned
  • Features of motion in an object
  • Causes and effects of motions

Hope it helps!

6 0
2 years ago
Read 2 more answers
You hold a metal block of mass 40 kg above your head at a height of 2 m.
Kitty [74]

Answer:

The work done by gravity is 784 J.

Explanation:

Given:

Mass of the block is, m=40\ kg

Height to which it is raised is, h=2\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,

\textrm{Work by gravity}=F_g\times h

Force of gravity is given as the product of mass and acceleration due to gravity.

\therefore F_g=mg=40\times 9.8=392\ N. Now,

\textrm{Work by gravity}=F_g\times h=392\times 2=784\ J

Therefore, the work done by gravity is 784 J.

5 0
3 years ago
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. On seeing the meter st
inessss [21]

Answer:

t=0.193s

Explanation:

What is said is that the meter fell d=18.3cm=0.183m under the action of gravity. We can use the formula for accelerated motion:

d=v_0t+\frac{at^2}{2}

Since it departed from rest it will mean that:

d=\frac{at^2}{2}

So our time will be:

t=\sqrt{\frac{2d}{a}}

Which for our values is:

t=\sqrt{\frac{2(0.183m)}{(9.81m/s^2)}}=0.193s

7 0
3 years ago
A supersaturated solution is one which A. has less solute dissolved than the solution should hold at that temperature. B. has mo
Anna35 [415]

Answer:

Option C is correct

Explanation:

A supersaturated solution is one that has more solute dissolved than the solution should hold at that temperature.

Examples include carbonated water, sugar syrup, honey.

A solution of a chemical compound can be dissolved in heated water to prepare a supersaturated solution. A solution becomes supersaturated as its temperature is changed.

3 0
3 years ago
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